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Oh.. so it shud have been +2* enthalpy change of combustion? am I right?
Yes, I suppose so
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Oh.. so it shud have been +2* enthalpy change of combustion? am I right?
Ratio is 3:2somebody solve me this pleasee
All MOLECULES. Silicon dioxide, sodium chloride, and zinc have a giant lattice structure. They are NOT molecules.But don't all molecules have van der waals forces?
Oh.. so it shud have been +2* enthalpy change of combustion? am I right?
Initial: b, 0, 0Help please??
View attachment 54318
I think it might be because if the pH is 14, there is a very high concentration of hydroxide ions. This would lead to OH- ions having high chances of collision with the aluminum and barium nitrate. It's just a guess I'm not sure thoughhttp://onlineexamhelp.com/wp-content/uploads/2014/08/9701_s14_qp_11.pdf
Guys , in Q8 - we are choosing ph 7 becaz the products are basic in nature ?
but this reaction isnt equilibrium related to do the shifting :/
Which molecules, each with a linear carbon chain, can have an optically active isomer?
I - C3H6BrI
II - C3H4BrI
III - C3H6I2
IV - C3H4Br2
A. I and II only
B. I, II and III only
C. II and III only
D. I, II and IV only
answer is B, explain?
Carbon monoxide burns readily in oxygen to form carbon dioxide.
What can be deduced from this information?
1 The +4 oxidation state of carbon is more stable than the +2 state.
2 The standard enthalpy change of formation of carbon dioxide is more negative than that of
carbon monoxide.
3 The value of the equilibrium constant for the reaction, 2CO(g) + O2(g) 2CO2(g), is likely to
be high.
answer is A, i need to know why 3 is correct? explanation please!
What's the answer?Help please??
View attachment 54318
hey, why did u do left= right? beacuse the charges must be balanced on both sides of the eq?Ratio is 3:2
3Ti+ + 2V5+ -----> 3Ti3+ + 2V(?)+
Left = 3+2*5=13
Right=3*3+2*?=9+2*?
Left=right
13=9+2*?
?=2
Thankyou!Initial: b, 0, 0
Final: b-x, x/2, x/2
Total moles = b-x+x/2+x/2= b
Mole ratio: (b-x)/b, (x/2)/b, (x/2)/b
Partial pressure = p(b-x)/b, px/(2b) px/(2b)
Kp = px/(2b)*px/(2b) / (p(b-x)/b)^2 = D
Alternatively you can realise that we can use concentration and find out Kp expression without taking pressure or mole ratio to consideration, since you can see Kp will have no units as pressure cancels out
Yeshey, why did u do left= right? beacuse the charges must be balanced on both sides of the eq?
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